Math 6140 Notes. Spring Codimension One Phenomena. Definition: Examples: Properties:


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1 Math 6140 Notes. Spring Codimension One Phenomena. A property of the points of a variety X holds in codimension one if the locus of points for which the property fails to hold is contained in a closed subset Z X whose components all have codimension 2 or more. We ll see several examples of this: A normal variety X is nonsingular in codimension 1 (Proposition 11.2). If X is normal then a rational map Φ : X > P n is regular in codimension 1 (on X) (Proposition 11.3). If Y is nonsingular and Φ : X Y is a surjective birational regular map and an isomorphism in codim 1, then Φ is an isomorphism (Prop 11.4) Blowing up a point is a simple example of a surjective birational regular map which is not an isomorphism. We ll consider this in some detail. Definition: If X is any variety and Z X is a closed subvariety, then the stalk of O X along Z is the ring: O X,Z := O X (U) C(X) {U U Z } Examples: (a) O X,X = C(X) (b) If Z = x X, then O X,Z = O X,x the stalk from 9. Properties: This stalk shares many of the properties from 9. (i) If V X is open and V Z, then O X,Z = O V,V Z. (ii) Each O X,Z is a local ring with maximal ideal: m Z := {φ O X,Z φ(z) 0 (where defined)} (iii) For each Φ : X Y, let W = Φ(Z) Y. Then there is a pullback: Φ : O Y,W O X,Z with Φ m W m Z (iv) If X is affine, let I(Z) C[X] be the prime ideal of Z. Then: O X,Z = C[X] I(Z) C(X) Thus by the correspondence between prime ideals C[X] I(Z) and prime ideals in C[X] contained in I(Z) ( 7) there is an inclusionreversing bijection: {prime ideals P O X,Z } {closed subvarieties W such that Z W X} and in particular 0 X and m Z Z. 1
2 So if cod X (Z) = 1, then m Z O X,Z is the unique (nonzero) prime ideal! As an example of rings with one nonzero prime ideal, consider: Definition: If K is a field, a function ν : K Z is a discrete valuation if: (i) ν(ab) =ν(a) +ν(b) and (ii) ν(a + b) min(ν(a),ν(b)) for all a, b K.Ifνis not the trivial (zero) valuation, then: A ν := {a K ν(a) 0} {0} is the discrete valuation ring (or DVR) associated to the valuation ν. Examples: (a) Fix a prime p. Fora Z {0}, define ν(a) = the largest power of p dividing a and ν( a)=ν(a) ν(b) for a b b Q. This is a discrete valuation and A ν = Z p (b) Fix a complex number c C, and for nonzero φ C(x) set: ν(φ) = the order of zero or pole (counted negatively) of φ at x This is a discrete valuation and A ν = C[x] x c (c) Again, for nonzero rational functions φ = f g C(x), let: ν(φ) = deg(g) deg(f) This is a discrete valuation and A ν = C[x 1 ] x 1. Observations: (a) If a valuation isn t surjective, its image is dz for some d>0, so we may as well divide through by d to get a surjective valuation. (b) In a DVR A ν (with surjective valuation ν), the (nonzero!) ideal: m := {a A ν ν(a) > 0} A ν is principal, generated by any element π A ν satisfying ν(π) = 1. Such an element is called a uniformizing parameter. Note that m = π is the unique (nonzero) prime ideal since everdeal in A ν is one of: m n = π n = {a A ν(a) n} for a uniquely determined n. Thus, in particular a DVR is always Noetherian and is a UFD (the factorization is b = uπ n for unit u and unique n). 2
3 Proposition 11.1: Suppose A is a (local) Noetherian domain with a unique nonzero prime ideal m A. Then: (a) For any f m, the ring A f is the field of fractions K = K(A). (b) If I A is any (nonzero) proper ideal, then I = m. (c) For each proper ideal I A, there is a unique n>0 such that: m n I but m n 1 I (d) If A is also integrally closed, then A is a DVR. Proof: (a) A f is a domain, and its prime ideals correspond to the prime ideals in A that do not contain f. Only the zero ideal has this property, so A f has no nonzero primes, hence the zero ideal is maximal and A f = K. (b) Suppose f m is arbitrary, and b A. Then b 1 = a K by (a), f n so ab = f n for some n. In other words, f b for all f m and b A. So m I for all ideals I and then m = I unless I = I = A. (c) It suffices to show that m n I for some n. But this follows from (b). That is, if a 1,..., a k = m = I (A is Noetherian!), then there are n 1,..., n k such that each a n i i I, and then we may let n =1+ (n i 1) to ensure that m n I. (d) Assume A is integrally closed. We ll first prove that m is principal. Pick an a m, and find n from (c) so that m n a but m n 1 a. Now choose b m n 1 a and consider φ = b K (the field of fractions of A). a By the choice of a and b, we know that φ A. On the other hand, φm A since b m n 1 and m n a. But φm m, since if it were, we d have: A[φ] m[φ] =m; α i φ i aα i φ i an inclusion of Amodules and since A is integrally closed, A[φ] is not finitely generated, so this would violate Noetherianness of A. Thus there is a unit u A and π m such that φπ = u. But now for any a m, wehave φa = a A, soa = a φ 1 =(a u 1 )π. Som = π. Now let s define the valuation. If a A, then either a is a unit or a m. In the latter case, we can write a = πa 1, and repeat the question of a 1. This gives an ascending chain of ideals: a a 1 a 2... that eventually stabilizes since A is Noetherian. 3
4 But this can only happen if a n = A (since π 1 A). Thus a n is a unit, and we can write: a = π n a n = uπ n and the value of n in this expression is evidently unique. Thus we may define: ν(uπ n )=n and extend to a valuation on K with the desired A = A ν. Proposition 11.2: A normal variets nonsingular in codimension 1. Proof: Since the nonsingular points of any variety are open (Prop 10.1) it suffices to show that given a normal variety X, there is no codimension 1 closed subvariety Z X such that Z Sing(X). For this, we may assume X is affine, so C[X] is integrally closed and then for any closed subvariety Z X: O X,Z = C[X] I(Z) C(X) is also integrally closed (see Proposition 9.4). When Z has codimension 1, this is a Noetherian ring with a unique prime, so by Proposition 11.1, the ring O X,Z is a DVR. In that case, let π m Z be a uniformizing parameter. Then π O X (U) for some U Z, and by shrinking U if necessary, we obtain an open subset V X such that: V = U f X is a basic open set (hence an affine variety) V Z I(V Z) = π C[V ] So let s replace X by V and Z by Z V. Here s the main point. If z Z and the images of f 1,..., f m C[X] under the map C[X] C[Z] O Z,z are a basis for the Zariski cotangent space Tz Z = m z/m 2 z of Z at z, then because I(Z) = π it follows that O Z,z = O X,z /π and so the images of f 1,..., f m,π under C[X] O X,z also span the Zariski tangent space Tz X of X at z. In other words, the dimensions of the Zariski tangent spaces satisfy: dim(o X,z ) dim(o Z,z )+1 But if z Z is a nonsingular point of Z, then dim(o Z,z ) = dim(z) and then the inequality above is an equality since dim(x) = dim(z) + 1and z X is also a nonsingular point of X. Since there is a nonempty open set of nonsingular points in Z, it follows that Z Sing(X), as desired. 4
5 Proposition 11.3: If Φ : X > CP n is a rational map and X is normal, then Φ is regular in codimension 1. Proof: Recall that if X CP m then in a neighborhood of of each point p X in the domain of X we can write Φ = (F 0 :... : F n ) for some d and F i C[x 0,..., x m ] d and conversely, so in particular, the domain of Φ is open. Thus to prove Φ is regular in codimension 1, we only have to prove that Φ is regular at some point p Z of each codimension 1 closed subvariety Z X. By Proposition 11.1 each O X,Z = C(X) ν for some valuation ν on C(X). Consider now the valuations of the ratios: ν( F i ) C(X) for some expression Φ = (F 0 :... : F n ) F j valid near some point of X. If we choose ν( F i 0 F j0 ) minimal (certainly negative!), then since they satisfy: ν( F i 0 F j0 )+ν( F i F i0 )=ν( F i F j0 ) ν( F i 0 F j0 ), each ν( F i F i0 ) 0. Thus the components of the rational map: Φ:X > C n = Ui0 CP n ; p ( F 0,..., F n ) F i0 F i0 are all in O X,Z, so for some open subset U X with U Z, each component belongs to U, and then Φ is regular all along U, as desired. Example: Consider the map to the line through the origin: π : C 2 > CP 1 ;(x, y) (x : y) As we ve seen, this cannot be extended to a regular map across (0, 0). But if we restrict this to the (nonsingular) curve C = V (y 2 x(x 1)(x λ)): π C : C CP 1 becomes regular across (0, 0). That s because (as in the Proposition): y 2 x = (x 1)(x λ) in a neighborhood of (0, 0), and then in that neighborhood: y 2 π(x, y) =( (x 1)(x λ) : y) =( y (x 1)(x λ) :1) so π(0, 0) = (0 : 1) (the vertical tangent line!). Notice that we couldn t do this if λ = 0, but in that case C is singular at (0, 0). 5
6 Remark: Putting Propositions 11.2 and 11.3 together in dimension 1 gives a very striking result. Namely, if X CP n is any quasiprojective variety of dimension 1 (i.e. a curve), let C be the normalization of the closure of X. Then C is a nonsingular (Proposition 11.2) projective curve with: C(C) = C(X) Moreover, if C is any other nonsingular projective curve with: C(C ) = C(C) then the associated birational map: Φ:C >C is an isomorphism(!) (by Proposition 11.3 applied to Φ and Φ 1 ). Thus up to isomorphism, there is only one nonsingular projective curve with a given field of rational functions. Moreover, if: C(C) C(C ) is an inclusion of such fields, then again by Proposition 11.3, the associated rational map: Φ:C C is actually regular, so the field inclusions correspond to (finite) maps of the nonsingular projective models of the fields. We will study more properties of nonsingular projective curves later. Proposition 11.4: If Φ : X Y is a birational surjective regular map and Y is nonsingular, then if Φ is not an isomorphism, it is also not an isomorphism in codimension 1. Proof: If Φ is not an isomorphism, let q Y be a point where Φ 1 is not defined and let p Φ 1 (q) (if Φ 1 were defined everywhere, then it would be a regular inverse of Φ!). Consider affine neighborhoods q V and p U Φ 1 (V ), and let U C n be a closed embedding. Then locally: Φ 1 =(φ 1,..., φ n ):V >U C n and to say that Φ 1 (q) is not defined is to say that some φ = φ i O Y,q. But we ve proved that O Y,q is a UFD in 10, so we can write φ = f in lowest g terms, with f,g O Y,q. When we pull back: Φ (φ) =x i C[U] istheith coordinate function, so: Φ (f) =x i Φ (g) O X,p and we can shrink V (and U) further so that f,g C[V ]. 6
7 Consider the hypersurface V (Φ (g)) U. We know p V (Φ (g)) since otherwise g(q) 0 would give f O g Y,q. Each component Z V (Φ (g)) (containing p) has codimension 1 in U by Krull s Principal Ideal Theorem. On the other hand, the image: Φ(Z) V (f) V (g) must have codimension 2inV. That s because if we write: g = ug 1 g n as a product of irreducibles (and shrink V again so that each g i C[V ]) then the V (g i ) are the irreducible components of V (g) and f (g i )=I(V(g i )) so V (f) V (g i ) V (g i ), and V (Z) is contained in one of these. Finally, by the theorem on the dimension of the fibers of a regular map, we see that since dim(z) > dim(φ(z)), it follows that every z Z is contained in a fiber of Φ of dimension 1, so Φ is not an isomorphism anywhere on Z. We ve actually shown that the set of points p X where Φ fails to be an isomorphism is a union of codimension 1 subvarieties of X. This is called the exceptional locus of the birational map Φ. Moral Example: It s time you saw the Cremona transformation: Φ:CP 2 > CP 2 ;(x : y : z) ( 1 x : 1 y : 1 )=(yz : xz : xy) x By Proposition 11.3, this is regular in codimension 1, and indeed it is: domain(φ) = CP 2 {(0:0:1), (0:1:0), (1 : 0 : 0)} Note that Φ 2 = id, but the exceptional locus of Φ are V (x) V (y) V (z) (this is only morally an example, since Φ isn t regular and it isn t surjective). Main Example: If X is nonsingular, then any rational map: Φ:X > CP n gives rise to a birational, surjective regular projection σ = π 1 :Γ Φ X from the closure of the graph of Φ (in X CP n )tox. Evidently, σ restricts to an isomorphism from Γ Φ to the domain of Φ, and the Proposition tells us that the exceptional set of σ is a union of codimsion 1 subvarieties, which is to say that the boundary Γ Φ Γ Φ is a union of codimension 1 subvarieties mapping to the indeterminate locus X dom(φ) of the rational map Φ. 7
8 The simplest instance of this is: The Blow up of a point in C n : Consider the rational map: Φ:C n > CP n 1 ; Φ(p 1,..., p n )=(p 1 :... : p n ) In other words, Φ maps a point p C n (other than the origin) to the point in projective space corresponding to the line through p and the origin. Then: bl 0 (C n ) := Γ Φ =Γ Φ ({0} CP n 1 ) because σ 1 (0) ({0} CP n 1 ) is required to have codimension 1 in bl 0 (C n ) (which has dimension n). This exceptional variety nicely fits with Γ Φ : Proposition 11.5: bl 0 (C n ) C n CP n 1 is: (a) nonsingular and (b) given by explicit equations (which we give in the proof). (σ is sometimes called a blowdown, σprocess or monoidal transformation) Proof: Consider the ring C[x 1,..., x n,y 1,..., y n ] graded by degree in y, as in the proof of Proposition 6.7. We ve seen that the closed subsets of C n CP n 1 are all of the form V (I) for homogeneous ideals I. Then: bl 0 (C n )=V := V ( x i y j x j ) since this contains Γ Φ (and no other points over 0 p C n ) and is closed and irreducible. Note that V does contain 0 CP n 1, as we proved it must(!) We see that V is irreducible and nonsingular by covering it with open affine subsets. Namely, let U i = C n (CP n 1 V ( )). Then: C[U i ]/I(V U i ) = C[x 1,..., x n, y 1,..., y n ]/ x i ( y j ) x j,..., x k ( y j ) x j ( y k ) y but all relations follow from the first ones, solving x j = x j i so this ring is a domain, V U i U i is closed and irreducible, and: C[U i ]/I(V U i )=C[V U i ] = C[x i, y 1,..., y n ] so that bl 0 (C n ) U i = V U i = C n is, in particular, nonsingular! Remark: bl 0 (C n ) defined above is not an affine variety, since there is a closed embedding CP n 1 bl 0 (C n ). It is quasiprojective, of course, as it sits inside C n CP n 1 as a (nonsingular) closed subvariety. 8
9 First Generalization: Blowing up a point in CP n. Consider π : CP n > CP n 1 the projection from a point p CP n. With suitable basis, we can assume p =0 C n CP n. It follows that: bl p (CP n ) := Γ π CP n CP n 1 is a nonsingular projective variety, with open (not affine!) cover given by: bl p (CP n )=bl 0 (C n ) (CP n p) and these open sets patch by the projection σ :Γ π CP n p. That was the easy generalization. Next: Proposition 11.6: If X CP n is an embedded variety (maybe singular) of dimension m and p X is a (maybe singular) point, then the variety: bl p (X) := Γ π X =Γ π X PC p X bl p (CP n ) where PC p X {p} CP n 1 is the projectivized tangent cone of X at p. If p is nonsingular then PC p (X) CP n 1 is naturalldentified with the projective tangent space to X at p, and bl p (X) is nonsingular along PC p (X). Proof: Let Y = X C n for p =0 C n CP n as above. It follows from Proposition 11.5 that the blowup of Y at p is the unique component: bl p (Y ) V ( x i y j x j,f(x 1,..., x n )) f I(Y ) ) containing the graph of π Y : Y > CP n 1. This is covered by: bl p (Y ) U i V ( f(x i y 1,..., x i y n ) ) C n = bl p (C n ) U i Expand f as a sum of homogeneous polynomials f = f d0 + f d f d. with f d0 0. Since 0 Y we have d 0 1 and then: gives: f(x i y 1,..., x i y n )=x d 0 i (f d0 ( y 1,..., y n )+x i f d0 ( y 1,..., y n )+...) V ( f(x i y 1,..., x i x n ) ) =V (x i ) V ( f(x y 1 y i,..., x n i x i x d 0 i ) ) 9
10 Consider the inclusion of polynomial rings (with z j = y j for convenience): A = C[x i z 1,..., x i z i 1,x i,x i z i+1,..., x i z n ] C[z 1,..., z i 1,x i,z i+1,..., z n ]=B The ideal P = {f(x i z 1,..., x i z n ) f I(Y )} A is prime by assumption, and the ideal Q = 1 f(x x d 0 i z 1,..., x i z n ) f I(Y ) B is equal to P xi B i for the natural inclusion B A xi. So it is prime, too. Since V (x i )=π1 1 (0) C n CP n 1 and V (Q) is irreducible, we get: bl p (Y ) U i = V (Q) and σ 1 (0) U i = V (Q) V (x i )=V( f d0 ( y 1,..., y n ) ) Thus σ 1 (0) = V ( f d0 (y 1,..., y n ) ) =:PC p X CP n 1 is the projectivized tangent cone to Y (or X) atp. To repeat, the f d0 (x 1,..., x n ) are the leading terms in the Taylor polynomials at p of the equations of Y C n. For the last part, notice that if F I(X) and f = F (1,x 1,..., x n ) I(Y ) then f 1 = n i=1 f x i (p)x i = n i=1 F x i (p)x i (up to a scalar) But n F n F V ( (p)x i )=π(v( (p)x i )) i=0 x i i=0 x i from which it follows that PC p (X) π(θ p ) (see Exercise 10.5). But if p is a nonsingular point, then π(θ p )) = CP m 1 so PC p (X) =π(θ p ) since they have the same dimension! Finally, if p is nonsingular, then: x i = x 1,..., x n,f 1 ( y 1,..., y n ) f I(Y ) = I(PC p (X) U i ) so the ideal of the nonsingular subvariety PC p (X) U i bl p (Y ) U i is generated by x i, and then as in the proof of Proposition 11.1, it follows that bl p (Y ) is nonsingular along PC p (X). Examples: (a) If Y = V (y 2 x 2 + x 3 ) C 2 is the nodal cubic, then: bl 0 (Y )=V(y x y ( ) 2 2 y2, 1+x) V (x y y ( ) 2 ( ) 3 1 y1 y1, 1 + y ) y 1 y 1 y 2 y 2 y 2 is nonsingular (the first open set is a parabola!), and: PC p (Y )=V (y 2 x 2 )=(0:1) (1 : 0) CP 1 consists of two points 10
11 (b) If Y = V (y 2 x 5 ), then: bl 0 (Y )=V (y x y 2 y 1, ( y2 and this is singular at the exceptional locus: y 1 ) 2 x 3 ) V (x y y 1 y 2, 1 y 3 PC p (Y )=V (y 2 )=(1:0) CP 1 ( ) 5 y1 ) Notice that PC p (Y ) is a single point (evidently nonsingular) but bl 0 (Y )is singular at this point. This is no contradiction since the ideal of the point is not given by a single equation! Final Remark: We can view the blowup bl p (X) as a surgery, removing the point p X and replacing it with the codimension 1 projectivized tangent cone PC p (X) bl p (X). We can ask whether blowing up X along an isolated singular point p X improves X, in the sense that bl p (X) is less singular than X was. The situation is somewhat complicated in dimension 3, but in dimension 2, the answer is yes. That is, if we start with an arbitrary projective surface X and normalize, then by Proposition 11.2, the resulting normal surface S has only finitely many (isolated) singularities, and then blowing up these singularities (and any singularities that occur in the exceptional loci) will eventually produce a nonsingular projective surface S n : S n = bl pn (S n 1 ) S n 1 = bl pn 1 (S n 2 ) S 1 = bl p1 (S) S Similarly, if Φ : S >S is a rational map of normal projective surfaces, then by Proposition 11.3, the indeterminacy locus of Φ is a finite set of points, and it is also true that after finitely many blowups, we obtain a regular map: S n = bl pn (S n 1 ). y 2 S 1 = bl p1 (S) Φ:S > S Thus with the assistance of blowups at points, the situation is somewhat similar (but more complicated) than the situation for curves. 11
12 Exercises If X is a normal variety, let: Div(X) = cod 1 irred V X Z[Z] i.e. Div(X) is the free abelian group generated by the codimension 1 closed subvarieties of X. Prove that there is a welldefined homomorphism: div : C(X) Div(X); div(φ) = V ν Z (φ) where ν Z is the discrete valuation defining O X,Z. Identify the kernel of div and prove that if X is an affine variety, then: C[X] is a UFD div is surjective (the quotient Div(X)/div(C(X) ) is called the class group of X). 2. Consider the projection from p =(1:0:0:0): π : CP 1 CP 1 = V (xw yz) > CP 2 ;(a : b : c : d) (b : c : d) Prove that bl p (CP 1 CP 1 ) is isomorphic to the compound blowup: X := bl σ 1 (0:1:0)(bl (1:0:0) (CP 2 )) (this is the blowup of CP 2 at the two points (0 : 1 : 0) and (1 : 0 : 0)). 3. If X = C(V ) PC n is the projective cone over V CP n 1, show that PC p (X) =V where p =0 C n CP n is the vertex of the cone and show that: π 2 : bl p (X) CP n 1 has fibers isomorphic to CP 1. In fact, find isomorphisms: bl p (X) U i = CP 1 (X V (x i )) 12
This is a closed subset of X Y, by Proposition 6.5(b), since it is equal to the inverse image of the diagonal under the regular map:
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